## Force and Power

One of the basic equations of physics isorIf the work is done in a timethen we can divide both sides byto obtain

**or**whereand

We can generalise this equation: If a body is travelling at a velocityagainst a forcethen the power needed to overcome the force isIt should be noted that at any instant the body may be accelerating or decelerating, so that v is an instantaneous, not necessarily a constant, speed. If a force is applied greater than the force needed to over come the resistive forcethen the body may accelerate, or the potential energy of the body may increase in some way. Given a powerand a velocitywe may find the driving force also using

Example

A lorry of mass 1500 kg moves along a straight horizontal road. The resistance to the motion of the

lorry has magnitude 750 N and the lorry’s engine is working at a rate of 36 kW.

a) Find the acceleration of the lorry when its speed is 20 m/s.

The lorry comes to a hill inclined at an angleto the horizontal, where

The magnitude of the resistance to motion from non-gravitational forces remains 750 N.

The lorry moves up the hill at a constant speed of 20 m/s.

b) Find the rate at which the lorry's engine is now working.

a)

The resultant force=1800-750=1050N

Now usefor the lorry – note thathere is the resultant force and not the 750N force in the question.

b) The lorry does work to overcome the force of friction and to increase the gravitational potenital energy by travelling up the hill.

In 1 second the lorry travels a distance v up the hill so from the diagram above, rises a vertical distance